Given the hypothesis "if x is odd or y is even". What are the cases? I want to prove an implication by discussing all the cases.
So if the hypothesis is "if x is odd or y is even", what are the cases?
I assume there are two cases:
1) x is odd and y is odd
2) x is even and y is even
But I'm wondering if there is a third case that "x is odd and y is even"?
I know "or" is used in the hypothesis, 
but in Mathematics, we say (A ∨ B) is true if either A or B is true.
And (A ∨ B) is also true if both A and B is true. 
In this case, if A is "x is odd" and B is "y is even", can I say "x is odd and y is even" is one of the cases?
But there is something more...
For example,
"For all integers x and y, if x is even and y is odd, then x+y is odd".
This statement is true. So its contrapositive "For all integers x and y, if x+y is even, then x is odd or y is even" is also true. However, in this case, the conclusion "x is odd or y is even" doesn't include the case "x is odd and y is even", otherwise the statement would be wrong.
This is what is confusing me.
Could anyone please help me to clarify this?
 A: 
If the hypothesis is "if x is odd or y is even", what are the cases?

You want to prove: If x is odd or y is even, then P(x,y) is true.
If you want to prove this by cases, your hypothesis gives you two cases to consider: (1) x is odd, (2) y is even. For each of these cases, however, there may be other cases sub-cases, sub-sub-cases, etc. to consider.
Sample Outline of a Proof by Cases
Suppose: x is odd or y is even (your hypothesis).

Case 1: Suppose x is odd.
(You might now consider two sub-cases, say: (1.1) y is odd, (1.2) y is even.)
Sub-case 1.1 : Assume y is odd. 
Prove: P(x,y) is true.
Sub-case 1.2: Assume y is even.
Prove: P(x,y) is true.
Conclude: If x is odd, then P(x,y) is true.

Case 2: Suppose y is even.
(You might now consider two sub-cases, say: (2.1) x is odd, (2.2) x is even.)
Sub-case 2.1: Assume x is odd. 
Prove: P(x,y) is true.
Sub-case 2.2: Assume x is even.
Prove: P(x,y) is true.
Conclude: If y is even, then P(x,y) is true.

(In both cases, we have proven that P(x,y) is true.)
As required: If x is odd or y is even, then P(x,y) is true.
